Question

A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate of 79.0 A/s. A circular coil with a single turn and radius rc = 20.0 cm encircles the solenoid with its plane perpendicular to the axis of the solenoid. Determine the magnitude of the average induced electric field in the coil.

Answer 1

E = 58.7 V/m

Step-by-step explanation:

As we know that flux linked with the coil is given as

`\phi = NBA`

here we have

`A = \pi R_s^2`

`B = \mu_o N i/L`

now we have

`\phi = N(\mu_o N i/L)(\pi R_s^2)`

now the induced EMF is rate of change in magnetic flux

`EMF = (d\phi)/(dt) = \mu_o N^2 \pi R_s^2 (di)/(dt)/L`

now for induced electric field in the coil is linked with the EMF as

`\int E. dL = EMF`

`E(2\pi r_c) = \mu_o N^2 \pi R_s^2 (di)/(dt)/L`

`E = (\mu_o N^2 R_s^2 (di)/(dt))/(2 r_c L)`

`E = ((4\pi * 10^(-7))(6500^2)(0.14^2)(79))/(2(0.20)(3.50))`

`E = 58.7 V/m`